Reaction on the beam - mechanics:
Determine algebraically the reaction on the beam loaded as shown in the figure given below. Neglect the thickness and mass of beam.
![1277_Reaction on the beam.png](https://www.expertsmind.com/CMSImages/1277_Reaction%20on%20the%20beam.png)
Resolved all the forces in horizontal and vertical direction.
Let reaction at hinged that is, point A is RAH and RAV, and reaction at the roller support is RBV
Let ∑H & ∑V is the sum of horizontal and vertical component of the forces ,The supported beam is in equilibrium, thus
∑H = ∑H = 0
∑H = RAH - 20cos60° + 30cos45° - 40cos80° = 0
RAH = - 4.26 KN ...(i)
V = RAV - 10 - 20sin60° - 30sin45° - 40sin80° + RBV = 0
RAV + RBV =87.92 KN ...(ii)
Taking moment about A
10 × 2 + 20sin60° × 6 + 30sin45° × 13 - 40sin80° × 17 - RBV × 17= 0
RBV = 62.9 KN ...(iii)
Putting the value of RBV in the equation (ii)
RAV = 25.02 KN
Thus RAH = - 4.26KN, RAV = 25.02KN, RBV = 62.9KN .......ANS